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Working on the Maxwell field as a gauge theory, at some point the following derivative comes up:

$\frac{\partial(\partial_iA_0)}{\partial A_0}=0$

which must be, accordingly to the theory, zero.

My doubt is: why is that necessarily true? Is this a specific feature of the form of the fields in this case? Or, maybe, is always true because of some assumption that the field and its derivatives are independent variables?

I'm confused about this because mathematically this is trivially not true and the most obvious "one-dimensional" example is:

$f=e^x$

$\frac{\partial}{\partial f}\frac{\partial f}{\partial x}=1$

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    $\begingroup$ It's just the field analogon to $\frac{\partial \dot{q}}{\partial q} = 0$ in the non-field case. If you understand that, you should understand this as well. If you don't understand that, go back to that non-field case. $\endgroup$
    – ACuriousMind ♦
    Commented Apr 11, 2016 at 22:13
  • $\begingroup$ Possible duplicates: physics.stackexchange.com/q/885/2451 , physics.stackexchange.com/q/526011/2451 and links therein. $\endgroup$
    – Qmechanic ♦
    Commented Apr 11, 2016 at 22:14
  • $\begingroup$ I understand the arguments for the independence between the (functional or not) variable and its derivatives on "both" formalisms, but I can't see why they're suficient. Is it not possible, for example, that $C_{1}e^{C_{2}t}$ is a solution for $q(t)$ so that even with the free choice of the initial conditions $q$ and $\dot{q}$ are not independent? I know that this is an old well answered question, sorry if it's annoying. $\endgroup$
    – GaloisFan
    Commented Apr 11, 2016 at 22:34

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